# 10percent of red balls are added to 20percent blue balls and total balls 24 and 3times the no. of red balls exceeds the blue ball by20 find the no. red and blue balls

Let the no. of red balls be R and blue balls be B

So, 3R = 20 + B

$\frac{10}{100}*R+\frac{20}{100}B=24\phantom{\rule{0ex}{0ex}}R+2B=240\phantom{\rule{0ex}{0ex}}R=240-2B\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}So,3*(240-2B)=20+B\phantom{\rule{0ex}{0ex}}720-6B=20+B\phantom{\rule{0ex}{0ex}}700=7B\phantom{\rule{0ex}{0ex}}B=100\phantom{\rule{0ex}{0ex}}R=240-2*100=40$

Regards

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